One of the new events in the 2002 Winter Olympics was the sport of skeleton. Sta

Question

One of the new events in the 2002 Winter Olympics was the sport of skeleton. Starting at the top of a steep, icy track, a rider jumps onto a sled (known as a skeleton) and proceeds - belly down and head first - to slide down the track. The track has fifteen turns and drops 104 m in elevation from top to bottom. (a) In the absense of non-conservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed of the rider at the beginning of the run is relatively small and can be ignored. (b) In reality, the best riders reach the bottom with a speed of 35.8 m/s (about 80 mi/h). How much work is done on an 82.9-kg rider and skeleton by non-conservative forces?

Explanation / Answer

Hi,   Given the initial velocity is 0 and the total vertical height = h = 104m. (a)  At the top it will have P.E. and at the bottom it will have only K.E.       Hence from conservation principle we have K.E = P.E                    => 0.5 m v^2 = mgh
                   => v^2 = 2gh = 2 * 9.8 * 104 = 2038.4
                   => v = 45.15 m/s

(b) mgh = 82.9 * 9.8 * 104 = 84491.68 J

     0.5 m v^2 = 0.5 * 82.9 * 35.8^2 = 53123.98 J

   Hence work done by non-conservative forces = 84491.68 - 53123.98 = 31367.7 J
Hope this helps you.

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